题目链接 （找不到原题链接拿题解博客将就一下）

以下是我的想法：

预处理当前可以进行的C的次数最大值，然后整一个小根堆。从前往后搜索，如果堆没满就加进去，堆满了就比较堆顶元素和当前元素，考虑是否更换。最后答案就是堆中元素的和。但是这样做WA了三个点，不知道是哪里有问题（

代码

#include <queue>
#include <cstdio>
#include <vector>

#define min(a, b) (a < b ? a : b)

const int N = 2e5 + 5;

int n, sum, c[N], e[N], minn[N];

std :: priority_queue <int, std :: vector<int>, std :: greater<int> > q;

int main() {
	freopen("dragons.in", "r", stdin);
	freopen("dragons.out", "w", stdout);
	scanf("%d", &n);
	for(int i = 1, x; i <= n; i++) {
		c[i] = -1;
		char ch = getchar();
		while(ch != 'c' && ch != 'e') ch = getchar();
		scanf("%d", &x);
		if(ch == 'c') c[i] = x;
		if(ch == 'e') {
			if(x <= 0 && i != n) {
		        printf("-1");
		        fclose(stdin); fclose(stdout);
		        return 0;
			}
			if(i != n) e[i] = x - 1;
			//else e[i] = x;
		}
	}
	minn[n + 1] = n + 1;
	for(int i = n; i >= 1; i--) {
		if(e[i]) minn[i] = min(minn[i + 1], e[i]);
		else minn[i] = minn[i + 1];
	}
	for(int i = 1; i <= n; i++) {
		if(c[i] >= 0) {
			if(q.size() < minn[i]) q.push(c[i]);
			else if(c[i] > q.top()) q.pop(), q.push(c[i]);
		}
	}
	if(q.size() < e[n]) {
		printf("-1");
		fclose(stdin); fclose(stdout);
		return 0;
	}
	while(!q.empty()) {
		sum += q.top();
		q.pop();
	}
	printf("%d", sum);
	fclose(stdin); fclose(stdout);
	return 0;
}